如何在bash等几个子过程完成并返回退出code！= 0时，任何与子code结束！= 0？ [英] How to wait in bash for several subprocesses to finish and return exit code !=0 when any subprocess ends with code !=0?

问题描述

如何从该脚本产生了几个子过程bash脚本等待完成并返回退出code！= 0时，任何一个子进程与code结束！= 0？

How to wait in a bash script for several subprocesses spawned from that script to finish and return exit code !=0 when any of the subprocesses ends with code !=0 ?

简单的脚本：

#!/bin/bash
for i in `seq 0 9`; do
  doCalculations $i &
done
wait

以上脚本将等待所有10催生子进程，但它总会给退出状态0（见帮助等待）。我怎么能修改此脚本，所以它会发现催生子进程的退出状态，并返回退出code 1时任子进程与code结束！= 0？

The above script will wait for all 10 spawned subprocesses, but it will always give exit status 0 (see help wait). How can I modify this script so it will discover exit statuses of spawned subprocesses and return exit code 1 when any of subprocesses ends with code !=0?

有没有那个比收集子进程的PID任何更好的解决方案，以便与和退出状态等待他们？

Is there any better solution for that than collecting PIDs of the subprocesses, wait for them in order and sum exit statuses?

推荐答案

等也（可选）主罚PID的过程中等待，并与$！你在后台启动的最后一个命令的PID。
修改循环存储的PID的每个产生子过程分成数组，然后循环再次等待每个PID

wait also (optionally) takes the PID of the process to wait for, and with $! you get the PID of the last command launched in background. Modify the loop to store the PID of each spawned sub-process into an array, and then loop again waiting on each PID.

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